/usr/share/pyshared/sepolgen/util.py is in python-sepolgen 1.2.1-1.
This file is owned by root:root, with mode 0o644.
The actual contents of the file can be viewed below.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 | # Authors: Karl MacMillan <kmacmillan@mentalrootkit.com>
#
# Copyright (C) 2006 Red Hat
# see file 'COPYING' for use and warranty information
#
# This program is free software; you can redistribute it and/or
# modify it under the terms of the GNU General Public License as
# published by the Free Software Foundation; version 2 only
#
# This program is distributed in the hope that it will be useful,
# but WITHOUT ANY WARRANTY; without even the implied warranty of
# MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
# GNU General Public License for more details.
#
# You should have received a copy of the GNU General Public License
# along with this program; if not, write to the Free Software
# Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA 02111-1307 USA
#
class ConsoleProgressBar:
def __init__(self, out, steps=100, indicator='#'):
self.blocks = 0
self.current = 0
self.steps = steps
self.indicator = indicator
self.out = out
self.done = False
def start(self, message=None):
self.done = False
if message:
self.out.write('\n%s:\n' % message)
self.out.write('%--10---20---30---40---50---60---70---80---90--100\n')
def step(self, n=1):
self.current += n
old = self.blocks
self.blocks = int(round(self.current / float(self.steps) * 100) / 2)
if self.blocks > 50:
self.blocks = 50
new = self.blocks - old
self.out.write(self.indicator * new)
self.out.flush()
if self.blocks == 50 and not self.done:
self.done = True
self.out.write("\n")
def set_to_list(s):
l = []
l.extend(s)
return l
def first(s, sorted=False):
"""
Return the first element of a set.
It sometimes useful to return the first element from a set but,
because sets are not indexable, this is rather hard. This function
will return the first element from a set. If sorted is True, then
the set will first be sorted (making this an expensive operation).
Otherwise a random element will be returned (as sets are not ordered).
"""
if not len(s):
raise IndexError("empty containter")
if sorted:
l = set_to_list(s)
l.sort()
return l[0]
else:
for x in s:
return x
if __name__ == "__main__":
import sys
import time
p = ConsoleProgressBar(sys.stdout, steps=999)
p.start("computing pi")
for i in range(999):
p.step()
time.sleep(0.001)
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